Ncert Exercises & Solutions

2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [ 1/32 – 1/n2]

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

The wavelength of transition = 1285 nm
= 1285 × 10–9 m (Given)

$v = 3.29 x 10^{15} (\frac{1}{3^{2}} - \frac{1}{n^{2}})$

$Since v = \frac{c}{λ}$
$= \frac{3.0 x 10^{8} m / s}{1285 x 10^{- 9} m}$
$Now,$
$v = 2.33 x 10^{14} s^{- 1}$
$Substituting the value of ν in the given expression,$
$3.29 x 10^{15} (\frac{1}{9} - \frac{1}{n^{2}}) = 2.33 x 10^{14}$
$\frac{1}{9} - \frac{1}{n^{2}} = \frac{2.33 x 10^{14}}{3.29 x 10^{15}}$
$\frac{1}{9} - 0.7082 x 10^{- 1} = \frac{1}{n^{2}}$
$\Rightarrow \frac{1}{n^{2}} = 1.1 x 10^{- 1} - 0.7082 x 10^{- 1}$
$\frac{1}{n^{2}} = 4.029 x 10^{- 2}$
$n = \sqrt{\frac{1}{4.029 x 10^{- 2}}}$
$n = 4.98 n$
$\approx 5$

Hence, for the transition to be observed at 1285 nm, n = 5.

The spectrum lies in the infra-red region.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions